2^x=x^2
2 to the power X equals x to the power 2
I have found a solution i.e 2 instantly but I found many answers so I operated by log which confused me

I'm not sure how you found "many answers" - I believe there are 3.

For positive values of x you can do this:


x^2 = 2^ x \\
2 \ln x = x \ln 2 \\
\frac 1 x \ln x = \frac 1 2 \ln 2\\
\frac 1 x \ln x = \ln ( \sqrt 2)


If you plot \small {f(x) = (1/x) \ln x} you'll see that f(x) equals \ln (\sqrt 2) at only two points, those being x = 2 and x=4. To prove that there are the only two solutions - if you find the derivative of this function you'll see that it equals zero only at x = e, so there can be at most two points where it equals \sqrt 2 . Hence there are only two solutions for positive values of x.

For negative values of x we can let y = -x, and then a similar approach yields


\frac 1 y \ln y = - \ln(\sqrt 2)


which has a solution at around y = 0.7667, so x = -0.7667 is the third solution. As a check - below is a plot of y_1 = x^2 and y_2 = 2^x and you can see the three points where the functions cross.